One-half the sum of two consecutive odd integers is −12. Find the smallest integer.

Step-by-step explanation:

Let the larger integer = 2x + 1

Let the second integer = 2x - 1

1/2 (2x + 1 + 2x - 1) = - 12               Combine the left

1/2(4x) = - 12                                  Combine 1/2 (4x)

2x = - 12                                         Divide by 2

2x/2 = - 12/2

x = - 6

2x + 1 = 2*-6 + 1 = -11

2x - 1 = 2*-6 - 1  = - 13

Check

-11 + - 13  = - 24

1/2(-24) = -12                               which checks.

x = ±sqrt(10)

step-by-step explanation:

5 - 2x² = -15

subtract 5 from each side

5-5 - 2x² = -15-5

-2x^2   = -20

divide by -2 on each side

-2x^2   /-2= -20/-2

x^2 =10

take the square root on each side

sqrt(x^2 )=±sqrt(10)

x = ±sqrt(10)

check:

5 - 2x² = -15

+sqrt(10)

5 - 2(sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15

-sqrt(10)

5 - 2(-sqrt(10))² = -15

5 -2(10) = -15

5 -20 = -15

-15 = -15